# Gibbs free energy in chemical reaction

I know this kind of question has been asked many times but I think this is a bit different.

We have $G = H - TS$. Here I am talking about chemical reactions at constant pressure and temperature. We can write $\Delta G = Q - Q_\mathrm{rev} = \Delta H - T \Delta S$.

For an irreversible chemical reaction, for example an exothermic one, if $q = -10~\mathrm{kJ~mol^{-1}}$ (exothermic), then $Q_\mathrm{rev}$ could be $-9$, $-8$, $-7$, etc. The difference $(Q – Q_\mathrm{rev})$ would be negative ($-1$, $-2$, $-3$, etc.). This represents the portion of $Q$ that can be converted into non-PV work. The remainder of the heat is dispersed into the surrounding and is wasted. Right?

So here $Q$ as absolute value is more than $Q_\mathrm{rev}$, so we have more heat generated than a reversible chemical reaction that can be used as work ($Q = 8~\mathrm{kJ~mol^{-1}}$ and $Q_\mathrm{rev} = 2~\mathrm{kJ~mol^{-1}}$). So why is it said that a reversible reaction produces more useful heat? Again, I am talking about chemical reaction and not mechanical process.

• Can you cite a reference for the equation $\Delta G = Q - Q_\mathrm{rev}$? Nov 20, 2015 at 18:05
• google.com/… Actually I found it on google and it is from www.usna.edu (a doc file). It is logical that ΔG=Q−Qrev since ΔS = Qrev/T Nov 20, 2015 at 18:12
• Can you explain why you find it logical, given that $\Delta G$ and $\Delta S$ are functions of state, while Q is not a function of state? Can you also provide the exact web site address? Nov 20, 2015 at 20:55
• Hello, it is a doc file inside the website which is downloaded automatically when you press on the link. The original website is www.usna.edu (US navy academy) Nov 20, 2015 at 21:11
• You can search it on google type: optional18 + usna.edu Nov 20, 2015 at 21:21